Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

Gears & gear trains

Meshing gears: n₁·N₁ = n₂·N₂
Torque: T₂ = T₁·(N₂/N₁)·η
Power: P = T·ω, ω = 2π·n/60
Compound train: total ratio = product of the stage ratios

Torsion of shafts

Shear stress: τ = T·r/J
Solid shaft: J = πd⁴/32
Hollow shaft: J = π(D⁴ − d⁴)/32
Angle of twist: φ = TL/(GJ)
Power: P = T·ω = 2πnT/60

Method

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Axial stress & strain

Normal stress: σ = F/A
Strain: ε = ΔL/L
Hooke’s law: σ = E·ε
Elongation: ΔL = FL/(EA)
Thermal strain: ε_T = α·ΔT

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

Mechanical vibrations

Natural frequency: ω_n = √(k/m), f_n = ω_n/(2π)
From static deflection: ω_n = √(g/δ_st)
Critical damping: c_c = 2√(km) = 2mω_n
Damping ratio: ζ = c/c_c
Damped frequency: ω_d = ω_n√(1 − ζ²)
Logarithmic decrement: δ = (1/n)·ln(x₀/x_n), ζ = δ/√(4π² + δ²)
Springs in parallel: k = k₁ + k₂
Springs in series: 1/k = 1/k₁ + 1/k₂

2Practice problems

1.Gears & gear trainsMedium

A 5.5 kW motor running at 1500 rpm drives a pinion with 45 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 97%. Find the torque T₁ on the motor shaft.

P = 5.5 kWn₁ = 1500 rpmN₁ = 45N₂ = 50η = 97%

T₁ = N·m

2.Gears & gear trainsHard

An open belt connects a 200 mm driving pulley at 2880 rpm to a 450 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 200 mmn₁ = 2880 rpmd₂ = 450 mmC = 1200 mmμ = 0.35T₁ = 1000 N

P = kW

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 300 mm rotate 9.74° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 12 mmL = 300 mmφ = 9.74°

G = GPa

4.Torsion of shaftsHard

A hollow shaft must transmit 5 kW at 750 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

P = 5 kWN = 750 rpmd_i/d_o = 0.5τ_allow = 40 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = −70 MPa, σᵧ = −60 MPa and τₓᵧ = 60 MPa (tension positive). The element is rotated 35° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = −70 MPaσ_y = −60 MPaτ_xy = 60 MPaθ = 35°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 160 MPa, σᵧ = 50 MPa and τₓᵧ = 20 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 160 MPaσ_y = 50 MPaτ_xy = 20 MPaθ = 50°

τ_θ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 50 mm, E = 70 GPa). Both are 500 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø20 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 70 GPaL = 500 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 60 mm, E = 100 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 200 MPatube: Ø120/Ø60 mm, E = 100 GPa, σ_allow = 70 MPa

P_max = kN

9.Column bucklingMedium

A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 2.5 mE = 200 GPa (steel)d = 40 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 3 mE = 200 GPa (steel)d = 80 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 5 cycles later by a peak of 11.4 mm. Find the damping ratio ζ.

x₀ = 40 mmx₅ = 11.4 mmn = 5

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 5 kN/m and k₂ = 50 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency f_n of vertical vibration in Hz.

m = 25 kgk₁ = 5 kN/mk₂ = 50 kN/mk₃ = 40 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T₁ = 35.01 N·m

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.

Motor torque

ω₁ = 2π·1500/60 = 157.1 rad/s

T₁ = P/ω₁ = 5500/157.1 = 35.01 N·m

Answer

T₁ = 35.01 N·m

2.Gears & gear trains

Answer: P = 19.35 kW

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = e^μθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Angle of wrap

θ = 2.933 rad

Slack-side tension

e^μθ = e^1.027 = 2.791, T₂ = 1000/2.791 = 358.3 N

Power

P = (1000 − 358.3)·30.16 = 19.35 kW

Answer

P = 19.35 kW

3.Torsion of shafts

Answer: G = 26.01 GPa

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Polar moment and angle

J = π·12⁴/32 = 2036 mm⁴

φ = 9.74·π/180 = 0.17 rad

Shear modulus

T = 30 N·m = 30000 N·mm

G = TL/(Jφ) = 30000·300/(2036·0.17) = 26006 MPa

G = 26.01 GPa

Answer

G = 26.01 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d_o = 20.52 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 5000/78.54 = 63.66 N·m

Outer diameter from the allowable stress

T = 63.66 N·m = 63662 N·mm

1 − k⁴ = 1 − 0.5⁴ = 0.9375

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·63662/(π·40·0.9375)) = 20.52 mm

d_i = 0.5·20.52 = 10.26 mm

Answer

d_o = 20.52 mm

A solid shaft would need d = 20.09 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: σ_y′ = −119.7 MPa

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

Transformation equation for the y′ face

σ_y′ = −65 − (−5)·0.342 − 60·0.9397 = −119.7 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = −10.33 MPa, σ_x′ + σ_y′ = −130 = σ_x + σ_y

Answer

σ_y′ = −119.7 MPa

6.Mohr’s circle & stress transformation

Answer: τ_θ = −57.64 MPa

Transformation equation

τ_θ = −55·0.9848 + 20·−0.1736 = −57.64 MPa

Answer

τ_θ = −57.64 MPa

7.Axial stress & strain

Answer: δ = 0.09017 mm

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 70·10³·3063 = 2.772 × 10⁸ N

Shortening

δ = 50000·500/2.772 × 10⁸ = 0.09017 mm

Answer

δ = 0.09017 mm

8.Axial stress & strain

Answer: P_max = 637.7 kN

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·8482 = 9.111 × 10⁸ N

Governing strain

ε = min(0.001, 0.0007) = 0.0007 (brass)

Largest load

P_max = 0.0007·9.111 × 10⁸ = 637743 N = 637.7 kN

Answer

P_max = 637.7 kN

9.Column buckling

Answer: P_cr = 9.922 kN

Euler buckling: a slender column becomes unstable at P_cr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Effective length

KL = 2·2500 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·125664/5000² = 9922 N = 9.922 kN

Answer

P_cr = 9.922 kN

10.Column buckling

Answer: σ_cr = 87.73 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 1·3000 = 3000 mm

KL/r = 3000/20 = 150

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/150² = 87.73 MPa

Answer

σ_cr = 87.73 MPa

11.Mechanical vibrations

Answer: ζ = 0.03992

Free vibration of a mass–spring system: the undamped natural frequency is ω_n = √(k/m) in rad/s, and f_n = ω_n/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/c_c, with critical damping c_c = 2·√(k·m) = 2·m·ω_n. An underdamped system (ζ < 1) oscillates at ω_d = ω_n·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.

Logarithmic decrement

δ = (1/5)·ln(40/11.4) = 0.2511

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2511/√(39.48 + 0.06303) = 0.03992

Answer

ζ = 0.03992

12.Mechanical vibrations

Answer: f_n = 4.844 Hz

Equivalent stiffness

Only k₁ and k₂ are in parallel. That pair is in series with k₃, so combine them as reciprocals.

k₁₂ = 5 + 50 = 55 kN/m

k_eq = 55·40/(55 + 40) = 23.16 kN/m

Natural frequency

ω_n = √(23158/25) = 30.44 rad/s

f_n = 30.44/(2π) = 4.844 Hz

Answer

f_n = 4.844 Hz

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